# How to solve this problem in detailed steps?

If $f(z)$ is analytic, prove that $$\left(\frac{d^{2}}{dx^{2}}+\frac{d^{2}}{dy^{2}}\right)\left|f\left(z\right)\right|^{2}=4\left|f^{\prime}\left(z\right)\right|^{2}$$

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What is P.T.?... –  DonAntonio Feb 15 '13 at 3:00
if you know that $4\frac{\partial}{\partial z}\frac{\partial}{\partial \overline{z}} = \Delta$, then it is easy. –  Yimin Feb 15 '13 at 3:18
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Michael Greinecker Feb 15 '13 at 20:12

We will use the following differential operators $$\frac{\partial}{\partial z}=\frac{1}{2}\left( \frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\qquad \frac{\partial}{\partial \bar{z}}=\frac{1}{2}\left( \frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$$ which commute and, for $C^2$ functions, satisfy $$\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}=4\frac{\partial^2}{\partial \bar{z}\partial z}.$$ Recall Cauchy-Riemann's equation, which characterizes analyticity (holomorphy): $$\frac{\partial f}{\partial \bar{z}}=0 \quad\Leftrightarrow \quad\frac{\partial \bar{f}}{\partial z}=0.$$

First, we have $$\frac{\partial}{\partial z}f\bar{f}=\frac{\partial f}{\partial z}\bar{f}+f\frac{\partial \bar{f}}{\partial z}=\frac{\partial f}{\partial z}\bar{f}.$$ Thus $$\frac{\partial }{\partial \bar{z}}\frac{\partial }{\partial z}f\bar{f}=\frac{\partial^2 f}{\partial\bar{z}\partial z}\bar{f}+\frac{\partial f}{\partial z}\frac{\partial \bar{f}}{\partial \bar{z}}=\frac{\partial}{\partial z}\frac{\partial f}{\partial \bar{z}}\bar{f}+\frac{\partial f}{\partial z}\overline{\frac{\partial f}{\partial z}}=|f'(z)|^2.$$

Since $|f|^2=f\bar{f}$, we now get the formula you wanted.

Let me know if you need more details.

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The question was actually posted by me and I agree with the above solution, but I eventually found the solution in this method and I think this is correct and easily understandable.

$$LHS = \left(\frac{\partial ^{2}}{\partial{x^{2}}}+{\frac{\partial ^{2}}{\partial{y^{2}}}}\right)\left|f\left(z\right)\right|^{2}$$
$$= \frac{\partial ^{2}}{\partial{x^{2}}}{\left(u^2+v^2\right)}+{\frac{\partial ^{2}}{\partial{y^{2}}}}{\left(u^2+v^2\right)}$$
Consider $u^2$ & $v^2$ as functions of functions u & v and apply Chain rule $$= \frac{\partial}{\partial{x}}{\left(2u*u_x+2v*v_x\right)}+{\frac{\partial}{\partial{y}}}{\left(2u*u_y+2v*v_y\right)}$$ Apply the second differentiation using product rule, $$=2\left(uu_{xx}+u_x^2+vv_{xx}+v_x^2+uu_{yy}+u_y^2+vv_{yy}+v_y^2\right)$$ But $u_{xx}+u_{yy}=0$, $v_{xx}+v_{yy}=0$, $u_x=v_y$, & $u_y=-v_x$ $$\therefore LHS=2\left(u_x^2+v_x^2+u_y^2+v_y^2\right)$$ $$LHS=4\left(u_x^2+v_x^2\right)$$

We know that $f^{\prime}\left(z\right)=u_x+iv_x=v_y-iu_y$ Then $$RHS=4\left|f^{\prime}\left(z\right)\right|^2$$ $$=4\left(u_x^2+v_x^2\right)$$ $$=LHS$$ Hence it is proven

Thank you for your help and correct my mistakes in the solution if any.

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